Codeforces 957C二分

最新推荐文章于 2021-03-04 19:02:28 发布

做不完的ACM

最新推荐文章于 2021-03-04 19:02:28 发布

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分类专栏：二分弱鸡的日常 CF日常

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本文介绍了一个CodeForces平台上的题目解析，该题目涉及能量转换效率的计算及最大值寻找算法。通过分析不同状态的能量值，使用三水平激光方案进行能量转换，并在特定约束条件下寻找最优解。

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An atom of element X can exist in n distinct states with energies E₁ < E₂ < ... < E_n. Arkady wants to build a laser on this element, using a three-level scheme. Here is a simplified description of the scheme.

Three distinct states i, j and k are selected, where i < j < k. After that the following process happens:

initially the atom is in the state i,
we spend E_k - E_i energy to put the atom in the state k,
the atom emits a photon with useful energy E_k - E_j and changes its state to the state j,
the atom spontaneously changes its state to the state i, losing energy E_j - E_i,
the process repeats from step 1.

Let's define the energy conversion efficiency as $\text{[math]}$ , i. e. the ration between the useful energy of the photon and spent energy.

Due to some limitations, Arkady can only choose such three states that E_k - E_i ≤ U.

Help Arkady to find such the maximum possible energy conversion efficiency within the above constraints.

Input

The first line contains two integers n and U (3 ≤ n ≤ 10⁵, 1 ≤ U ≤ 10⁹) — the number of states and the maximum possible difference between E_k and E_i.

The second line contains a sequence of integers E₁, E₂, ..., E_n (1 ≤ E₁ < E₂... < E_n ≤ 10⁹). It is guaranteed that all E_i are given in increasing order.

Output

If it is not possible to choose three states that satisfy all constraints, print -1.

Otherwise, print one real number η — the maximum possible energy conversion efficiency. Your answer is considered correct its absolute or relative error does not exceed 10^- 9.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if $\text{[math]}$ .

Examples

Input

Copy

4 4
1 3 5 7

Output

0.5

Input

Copy

10 8
10 13 15 16 17 19 20 22 24 25

Output

0.875

Input

Copy

3 1
2 5 10

Output

-1

Note

In the first example choose states 1, 2 and 3, so that the energy conversion efficiency becomes equal to $\text{[math]}$ .

In the second example choose states 4, 5 and 9, so that the energy conversion efficiency becomes equal to $\text{[math]}$ .

题解：首先一定是相邻2个I,J数，然后尽可能使K最大化，所以可以N*LOGN，当然也有O(N)的做法，逆过来写，维护个不断减少的右指针。注意二分时候要upper_bound,然后-1。因为如果正好存在EJ+U的话，返回的是下一个，然后-1变成EJ+U。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long double ld;

#define x0 x0___
#define y0 y0___
#define pb push_back
#define SZ(X) ((int)X.size())
#define mp make_pair
#define fi first
#define se second
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pli pair<ll,int>
#define pil pair<int,ll>
#define ALL(X) X.begin(),X.end()
#define RALL(X) X.rbegin(),X.rend()
#define rep(i,j,k) for(int i = j;i <= k;i ++)
#define per(i,j,k) for(int i = j;i >= k;i --)
#define mem(a,p) memset(a,p,sizeof(a))


const ll MOD = 1E9 + 7;
ll qmod(ll a,ll b,ll c) {ll res=1;a%=c; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%c;a=a*a%c;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}

template<typename T, typename S>
void upmax(T& a,S b){if(a<b) a=b;}
template<typename T, typename S>
void upmin(T& a,S b){if(a>b) a=b;}
template<typename T>
void W(T b){cout << b << endl;}
void gettle() {while(1);}
void getre() {int t=0;t/=t;}


/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////



const int N = 1E5 + 7;
int a[N];
int main()
{
    int n, U;
    scanf("%d %d", &n, &U);
    rep(i,1,n) scanf("%d",&a[i]);
    double res = -1.0;
    rep (i,1,n-2) {
        int t1 = a[i];
        int t2 = a[i+1];
        int up = U + t1;
        int k = upper_bound(a+1, a+1+n, up) - a - 1;
        if(k < i + 2) continue;
        upmax(res, (a[k]-t2*1.0)/(a[k]-t1));
    }
    return !printf("%.15f\n", res);
}